# A projectile is shot from the ground at an angle of pi/12  and a speed of 7 /18 m/s. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

##### 3 Answers
Jul 19, 2017

distance from starting point is $\frac{5}{1296} m \approx 3.858 \text{x} {10}^{-} 3 m$

#### Explanation:

Range, $R = {V}^{2} / g \sin \left(2 \theta\right)$

When the particle reaches its maximum height, the distance from starting point will be $\frac{R}{2}$

$\frac{R}{2} = {V}^{2} / \left(2 g\right) \sin \left(2 \theta\right)$ where V is the initial launching speed

$\frac{R}{2} = {\left(\frac{7}{18}\right)}^{2} / \left(2 g\right) \sin \left(2 \cdot \frac{\pi}{12}\right)$

$\frac{R}{2} = \frac{\frac{49}{324}}{2 g} \sin \left(\frac{\pi}{6}\right)$

[Take $g = 9.8 m {s}^{-} 2$]

$\frac{R}{2} = \left(\frac{5}{648}\right) \sin \left(\frac{\pi}{6}\right)$

$\frac{R}{2} = \frac{5}{1296} m \approx 3.858 \text{x} {10}^{-} 3 m$

Jul 19, 2017

$\text{distance} = 0.00389$ $\text{m}$

#### Explanation:

We're asked to find the distance from the launch point a projectile is when it reaches its maximum height.

We can use the equation

${\left({v}_{y}\right)}^{2} = {\left({v}_{0 y}\right)}^{2} - 2 g h$

to find the maximum height $h$.

When the particle is at its maximum height, the instantaneous velocity ${v}_{y}$ is $0$.

The initial $y$-velocity ${v}_{0 y}$ is

${v}_{0 y} = {v}_{0} \sin {\alpha}_{0} = \left(\frac{7}{18} \textcolor{w h i t e}{l} \text{m/s}\right) \sin \left(\frac{\pi}{12}\right) = 0.101$ $\text{m/s}$

Plugging in known values, we have

$0 = \left(0.101 \textcolor{w h i t e}{l} {\text{m/s")^2 - 2(9.81color(white)(l)"m/s}}^{2}\right) h$

h = ((0.101color(white)(l)"m/s")^2)/(2(9.81color(white)(l)"m/s"^2)) = color(red)(5.16xx10^-4 color(red)("m"

To find the horizontal distance covered $\Delta x$, we use the equation

$\Delta x = {v}_{0 x} t$

${v}_{0 x} = {v}_{0} \cos {\alpha}_{0} = 0.376$ $\text{m/s}$

The time $t$ when it reaches its maximum height is given by

${v}_{y} = {v}_{0 y} - g t$

Rearranging this equation and plugging in known values:

$t = \frac{{v}_{0 y} - {v}_{y}}{g} = \left(0.101 \textcolor{w h i t e}{l} {\text{m/s" - 0)/(9.81color(white)(l)"m/s}}^{2}\right) = 0.0103$ $\text{s}$

Thus, we have

Deltax = (0.376color(white)(l)"m/s")(0.0103color(white)(l)"s") = color(green)(0.00385 color(green)("m"

The distance from the starting point is thus

$r = \sqrt{{\left(\Delta x\right)}^{2} + {h}^{2}} = \sqrt{{\left(\textcolor{g r e e n}{0.00385 \textcolor{w h i t e}{l} \text{m"))^2 + (color(red)(5.16xx10^-4color(white)(l)"m}}\right)}^{2}}$

= color(blue)(0.00389 color(blue)("m"

Jul 19, 2017

THe distance is $= 0.0039 m$

#### Explanation:

Resolving in the vertical direction ${\uparrow}^{+}$

initial velocity is ${u}_{y} = v \sin \theta = \frac{7}{18} \cdot \sin \left(\frac{1}{12} \pi\right)$

Acceleration is $a = - g$

At the maximum height, $v = 0$

We apply the equation of motion

$v = u + a t$

to calculate the time to reach the greatest height

$0 = \frac{7}{18} \sin \left(\frac{1}{12} \pi\right) - g \cdot t$

$t = \frac{7}{18 g} \cdot \sin \left(\frac{1}{12} \pi\right)$

$= 0.0103 s$

The greatest height is

$h = {\left(\frac{7}{18} \sin \left(\frac{1}{12} \pi\right)\right)}^{2} / \left(2 g\right) = 0.00052 m$

Resolving in the horizontal direction ${\rightarrow}^{+}$

To find the horizontal distance, we apply the equation of motion

$s = {u}_{x} \cdot t$

$= \frac{7}{18} \cos \left(\frac{1}{12} \pi\right) \cdot 0.0103$

$= 0.00387 m$

The distance from the starting point is

$d = \sqrt{{h}^{2} + {s}^{2}}$

$= \sqrt{{0.00052}^{2} + {0.00387}^{2}}$

$= 0.0039 m$